Let f: R → R satisfy the equation f ( x + y ) = f ( x ) · f ( y ) for all x, y ∈ R and f ( x ) ≠ 0 for any x ∈ R . If the function f is differentiable at x = 0 and f ^ ' ( 0 ) = 3 , then _ h → 0 (1)/(h) (f (h) - 1) is

Correct answer: 0 If f ( x + y ) = f ( x ) · f ( y ) and f ^ ' ( 0 ) = 3 then f ( x ) = a ^ x ⇒ f ^ ' (x) = a ^ x · lna ⇒ f ^ ' (0) = lna = 3 ⇒ a = e ^ 3 ⇒ f(x) = ( e ^ 3 ) ^ x = e ^ 3x _ x → 0 (f ( x ) - 1)/(x) = _ x → 0 ( e ^ 3x - 1 3x × 3) = 1 × 3 = 3

JEE Main 202118 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 18 March
Shift: Evening Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2021 · 18 March · Evening Shift · Mathematics · Continuity and Differentiability

Let f: R → R satisfy the equation f ( x + y ) = f ( x ) · f ( y )...

Let

f: R \to R

satisfy the equation

f ( x + y ) = f ( x ) \cdot f ( y )

for all

x, y \in R

and

f ( x ) \ne 0

for any

x \in R

. If the function

f

is differentiable at

x = 0

and

{f}^{'} ( 0 ) = 3

, then

\lim_{h \to 0}\dfrac{1}{h}

\left(f

\left(h\right) - 1\right)

is equal to ___ .

Correct answer

Step-by-step solution

If $f ( x + y ) = f ( x ) \cdot f ( y ) \text{ and } {f}^{'} ( 0 ) = 3$ then

$f ( x ) = {a}^{x} ⇒ {f}^{'}\left(x\right) = {a}^{x} \cdot lna$

$⇒ {f}^{'}\left(0\right) = lna = 3 ⇒ a = {e}^{3}$

$⇒ f\left(x\right) = {\left({e}^{3}\right)}^{x} = {e}^{3x}$

$\lim_{x \to 0}\dfrac{f ( x ) - 1}{x} = \lim_{x \to 0}\left(\dfrac{{e}^{3x} - 1}{3x} \times 3\right) = 1 \times 3 = 3$

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